That is, if x<0, then f(x)=−x, if 0≤x<3, then f(x)=3−x and if x>3, then f(x)=(x−3)2.

Theorem used:

1. The functions such as “Polynomials, rational functions, root functions, trigonometric functions, inverse trigonometric functions, exponential functions and logarithmic functions” are continuous at every number in their domains.

2. A function f is continuous from the right at a number a if limx→a+f(x)=f(a) and a function f is continuous from the left at a number a if limx→a−f(x)=f(a).

3. The limit limx→af(x)=L if and only if limx→a−f(x)=L=limx→a+f(x).

Section (i)

Obtain the limit of the function f(x) as x approaches right side of 0.

If 0≤x<3, then f(x)=3−x.

Here, f(x)=3−x is a polynomial defined in the interval [0,3) and by theorem 1, it is continuous everywhere on its domain [0,3). That is, limx→af(x)=f(a) for every a∈[0,3). Therefore, the limit exist.

Consider the right hand limit limx→0+f(x).

limx→0+f(x)=limx→0+(3−x)=f(0)[by theorem 2]=3−(0)=3

Thus, the limit of the function f(x)=3−x as x approaches right side of 0 is 3.

Section (ii)

Obtain the limit of the function f(x) as x approaches left side of 0.

If x<0, then f(x)=−x.

Here, f(x)=−x is a root function defined in the interval (−∞,0) and by theorem 1, it is continuous everywhere on its domain (−∞,0). That is, limx→af(x)=f(a) for every a∈(−∞,0). Therefore, the limit exist.

Consider the left hand limit limx→0−f(x).

limx→0−f(x)=limx→0−(−x)=−0[by theorem 2]=0

Thus, the limit of the function f(x)=−x as x approaches left side of 0 is 0